Where are the inaccuracies in math.sqrt() and math.pow() coming from for large numbers? [duplicate]

Question

If you take a number, take its square root, drop the decimal, and then raise it to the second power, the result should always be less than or equal to the original number.

This seems to hold true in python until you try it on 99999999999999975425 for some reason.

import math

def check(n):
    assert math.pow(math.floor(math.sqrt(n)), 2) <= n

check(99999999999999975424)  # No exception.
check(99999999999999975425)  # Throws AssertionError.

It looks like math.pow(math.floor(math.sqrt(99999999999999975425)), 2) returns 1e+20.

I assume this has something to do with the way we store values in python... something related to floating point arithmetic, but I can't reason about specifically how that affects this case.

Answer 1

Unlike Evan Rose's (now-deleted) answer claims, this is not due to an epsilon value in the sqrt algorithm.

Most math module functions cast their inputs to float, and math.sqrt is one of them.

99999999999999975425 cannot be represented as a float. For this input, the cast produces a float with exact numeric value 99999999999999983616, which repr shows as 9.999999999999998e+19:

>>> float(99999999999999975425)
9.999999999999998e+19
>>> int(_)
99999999999999983616L

The closest float to the square root of this number is 10000000000.0, and that's what math.sqrt returns.